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I am sure that it holds for ordered sets but in the case of unordered set such as $$A={(1,2,3)}, B={(5,1,4)}$$ it does not hold.
As $A+B$ will be ${(6,3,7)}$.

$\inf(A+B)=3$ $$\inf(A)+\inf(B)=1+1=2$$

AsukaMinato
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Avyansh Katiyar
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    The sum of sets consists of elements that are the sums of all possible pairs of elements of each of the sets, and not just the corresponding – thing Jan 05 '20 at 12:37
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    The problem is that you are misunderstanding "A+ B". Given sets A= {1, 2, 3} and B= {5, 1, 4}, then A+ B= {1+ 5. 1+ 1. 1+ 4. 2+ 5. 2+ 1, 2+ 4, 3+ 5, 3+ 1, 3+ 4}= {6, 2, 5, 7, 3, 8, 4}= {2, 3, 4, 5, 6, 7, 8}. (both 1+ 5 and 2+ 4 are 6 and both 2+ 5 and 3+ 4 are 7 but we do not write the same number twice.) – user247327 Jan 05 '20 at 12:43

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